Complex number fraction Find the real and imaginary number:
$$\frac{1}{z}=\frac{2}{2+j3}+\frac{1}{3-j2}$$
How do I invert $\frac{1}{z}$ to $z$ so that I can start solving it?
 A: 
Solve a more general way, assume $q\in\mathbb{C}$:
$$q=\Re[z]+\Im[z]i$$


So, we get:
$$\frac{1}{q}=\frac{\overline{q}}{q\overline{q}}=\frac{\overline{q}}{|q|^2}=\frac{\Re[q]-\Im[q]i}{\Re^2[q]+\Im^2[q]}=\frac{\Re[q]}{\Re^2[q]+\Im^2[q]}-\frac{\Im[q]}{\Re^2[q]+\Im^2[q]}\cdot i$$
Now, in your case:
$$\frac{2}{2+3i}+\frac{1}{3-2i}=\frac{2(3-2i)+2+3i}{(2+3i)(3-2i)}=\frac{8-i}{12+5i}=$$
$$\frac{(8-i)\cdot\overline{12+5i}}{(12+5i)\cdot\overline{12+5i}}=\frac{91-52i}{12^2+5^2}=\frac{91-52i}{169}=\frac{7-4i}{13}\to\color{red}{z=\frac{13}{7-4i}}$$
For $z$ we find:
$$z=\frac{13}{7-4i}=\frac{13\cdot\overline{7-4i}}{(7-4i)\cdot\overline{7-4i}}=\frac{13(7+4i)}{7^2+4^2}=\frac{7+4i}{5}$$
So:

*

*$$\Re[z]=\Re\left[\frac{7+4i}{5}\right]=\frac{7}{5}$$

*$$\Im[z]=\Im\left[\frac{7+4i}{5}\right]=\frac{4}{5}$$
A: First of all, get rid of the denominators. Since it has to be $z \neq 0$, we can multiply both sides by $z(2 + j3)(3 - j2)$, and get:
$$(2 + j3)(3 - j2) = 2z(3 - j2) + z(2 + j3)$$
With simple algebraic manipulations, we have
$$6 + j9 - j4 + 6 = 6z - j4z + 2z + j3z \implies 12 + j5 = (8 - j)z$$
That is,
$$z = \frac{12 + j5}{8 - j} = \frac{12 + j5}{8 - j}\cdot\frac{8 + j}{8 + j} = \ldots$$
A: after the hint above we get
$$\frac{1}{z}=\frac{4-6i}{13}+\frac{3+2i}{9+4}$$ thus
$$\frac{1}{z}=\frac{7-4i}{13}$$
thus $$z=\frac{13}{7-6i}$$
simplifying this we have $$z=\frac{13(7+6i)}{85}$$
A: The denominators of the right hand side are closely related.

We obtain
  \begin{align*}
\frac{1}{z}&=\frac{2}{2+j3}+\frac{1}{3-j2}=\frac{2}{2+j3}+\frac{j}{2+j3}
\end{align*}
  It follows
  \begin{align*}
z=\frac{2+j3}{2+j}=\frac{1}{5}(2+j3)(2-j)=\frac{1}{5}(7+j4)
\end{align*}

